CF 104255A - Stickers for BSUIR Open

We are given a rectangular sheet of paper with dimensions $n times m$. From this sheet, we want to cut out $k$ identical square stickers, where each sticker is a square with side length $x$. The squares must be fully contained in the sheet, and they are not allowed to overlap.

codeforcescompetitive-programming

CF 104255A - Stickers for BSUIR Open

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Solution

Problem Understanding

We are given a rectangular sheet of paper with dimensions $n \times m$. From this sheet, we want to cut out $k$ identical square stickers, where each sticker is a square with side length $x$. The squares must be fully contained in the sheet, and they are not allowed to overlap.

The task is to determine the maximum possible integer or real value of $x$ such that at least $k$ squares of size $x \times x$ can be placed inside the rectangle.

The key observation is that for any fixed side length $x$, the number of squares we can place is determined by how many fit along each dimension. Horizontally we can fit $\lfloor n / x \rfloor$ squares, and vertically we can fit $\lfloor m / x \rfloor$ squares, so the total is their product.

The constraints $n, m \le 10^3$ are small enough that evaluating this function is trivial for a fixed $x$, but $k \le 10^9$ makes it impossible to enumerate possibilities for placement counts directly. The answer is not necessarily an integer, since the optimal square size can be fractional.

A naive approach would try all possible square sizes from $1$ down to very small increments, but this fails because the answer has continuous precision up to $10^{-6}$. Even discretizing to $10^{-6}$ resolution would require about $10^6$ checks, and each check is $O(1)$, which is borderline but unnecessary given a cleaner structure.

Edge cases appear when $k$ is very large, forcing the square size to be small. For example, if $n = m = 2$ and $k = 3$, then $x = 1$ works because $2$ squares fit per row and column, giving $4$ total, but any $x > 1$ reduces the count below $3$. Another edge case is when $k = 1$, where the answer is simply $\min(n, m)$, since one square can occupy the largest possible side.

Approaches

The brute-force idea is to try all possible square side lengths $x$, compute how many squares fit, and track the largest valid value. If we discretize $x$ into small steps such as $10^{-6}$, each check requires computing $\lfloor n/x \rfloor \cdot \lfloor m/x \rfloor$, which is constant time. However, this discretization leads to about $10^6$ candidates, and each evaluation involves floating-point divisions and floor operations. While this might pass in some cases, it is unnecessary and still fragile with precision.

The key structural observation is that the number of squares that fit is monotonic with respect to $x$. If a given size $x$ works, then any smaller size also works because reducing the square size only increases how many fit along each dimension. Conversely, if $x$ is too large, increasing it further only reduces feasibility. This monotonic behavior allows us to treat the problem as a continuous binary search over $x$.

We define a predicate $f(x)$ that checks whether $\lfloor n/x \rfloor \cdot \lfloor m/x \rfloor \ge k$. This predicate is monotonic, so we can binary search the maximum $x$ for which it remains true.

The search space is from $0$ to $\max(n, m)$. Each step of binary search evaluates the predicate in constant time, giving a logarithmic number of iterations sufficient for $10^{-6}$ precision.

Approach Time Complexity Space Complexity Verdict
Brute Force $O(10^6)$ $O(1)$ Too slow / imprecise
Binary Search $O(\log(\text{precision}))$ $O(1)$ Accepted

Algorithm Walkthrough

We reformulate the problem as finding the largest $x$ such that at least $k$ squares of side $x$ fit.

  1. Define a function can(x) that computes how many squares fit: $(n // x) * (m // x)$. If this value is at least $k$, then $x$ is feasible. The floor division captures how many full squares fit along each axis.
  2. Set the binary search interval with left = 0 and right = max(n, m). The answer must lie in this range because a square larger than the rectangle cannot fit even once.
  3. Repeatedly compute mid = (left + right) / 2 and test can(mid).
  4. If can(mid) is true, we can try to increase the square size, so we move left = mid. Otherwise, we reduce the size with right = mid.
  5. Continue until the interval width is below $10^{-7}$, which guarantees correctness within the required error bound.
  6. Output left as the best feasible approximation.

Why it works

The function $can(x)$ is monotonic decreasing in $x$. Once a size becomes infeasible, all larger sizes remain infeasible because increasing $x$ can only decrease or maintain the number of squares that fit. This monotonicity guarantees that binary search never skips over the optimal boundary, and the final convergence point approximates the largest feasible $x$ within the required precision.

Python Solution

import sys
input = sys.stdin.readline

def can(n, m, k, x):
    return (n // x) * (m // x) >= k

def solve():
    n, m, k = map(int, input().split())

    # handle edge case: no squares fit unless x <= min(n, m)
    left, right = 0.0, float(max(n, m))

    for _ in range(60):
        mid = (left + right) / 2
        if mid == 0:
            left = mid
            continue
        if (n // mid) * (m // mid) >= k:
            left = mid
        else:
            right = mid

    print(left)

if __name__ == "__main__":
    solve()

The core of the implementation is the can logic, which directly encodes how many squares fit along each dimension. The binary search runs for a fixed number of iterations (60), which is more than enough for double precision convergence.

One subtle point is avoiding division by zero when mid becomes extremely small. In practice, the loop converges before that becomes an issue, but the check ensures numerical safety. Another subtlety is using 60 iterations instead of an epsilon condition, which avoids floating-point instability.

Worked Examples

Example 1

Input:

2 2 3

We expect a square size of $1.0$.

Step left right mid fit = (n//mid)*(m//mid) feasible
1 0.0 2.0 1.0 4 yes
2 1.0 2.0 1.5 1 no
3 1.0 1.5 1.25 1 no
4 1.0 1.25 1.125 1 no

The search converges toward 1.0 as the largest feasible value. Any value above 1 reduces the number of fit squares below 3.

Example 2

Input:

3 3 1

We expect the answer $3.0$.

Step left right mid fit feasible
1 0.0 3.0 1.5 4 yes
2 1.5 3.0 2.25 1 yes
3 2.25 3.0 2.625 1 yes

This shows that even large values remain feasible until we approach the true limit $3$, confirming that the upper bound is tight.

Complexity Analysis

Measure Complexity Explanation
Time $O(\log(1/\epsilon))$ Binary search over real values with constant-time feasibility check
Space $O(1)$ Only a few scalar variables used

The binary search converges in about 60 iterations, which is trivial under the 1 second limit even in Python.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from math import floor

    # inline solution for testing
    n, m, k = map(int, sys.stdin.readline().split())

    def can(x):
        if x == 0:
            return True
        return (n // x) * (m // x) >= k

    l, r = 0.0, float(max(n, m))
    for _ in range(60):
        mid = (l + r) / 2
        if mid == 0:
            break
        if (n // mid) * (m // mid) >= k:
            l = mid
        else:
            r = mid
    return str(l)

# provided sample
assert abs(float(run("2 2 3\n")) - 1.0) < 1e-6

# minimum case
assert abs(float(run("1 1 1\n")) - 1.0) < 1e-6

# large k forcing small answer
assert float(run("10 10 1000000000\n")) < 1e-6

# perfect tiling
assert abs(float(run("4 4 4\n")) - 2.0) < 1e-6

# asymmetric rectangle
assert abs(float(run("6 3 2\n")) - 3.0) < 1e-6
Test input Expected output What it validates
1 1 1 1.0 minimal valid case
10 10 1000000000 ~0 extreme density forcing tiny squares
4 4 4 2.0 exact tiling boundary
6 3 2 3.0 asymmetric packing

Edge Cases

When $k = 1$, the algorithm naturally expands toward the largest feasible square, which becomes $\min(n, m)$. The binary search starts at zero and quickly confirms feasibility for large mid values until it converges to the true maximum.

When $k$ is extremely large, the feasibility check becomes false for almost all mid values except those near zero. The search shrinks the interval until the square size becomes small enough that both floor divisions still produce enough placements.

When $n = m$, symmetry does not simplify the logic, but it makes the feasibility boundary sharper. The algorithm still converges correctly because monotonicity is preserved regardless of symmetry.